2017 AMC 10A Problem 23

Attempt Problem 23 of the 2017 AMC 10A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2017 AMC 10A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

23.

How many triangles with positive area have all their vertices at points (i,j)(i,j) in the coordinate plane, where ii and jj are integers between 11 and 5,5, inclusive?

21282128

21482148

21602160

22002200

23002300

Answer: B
Concepts:lattice pointcounting shapes in figurescomplementary counting
Difficulty rating: 2250
Solution:

We can use complementary counting to find the total number of triangles and subtract out the ones that don't work.

There are a total of 52=255^2 = 25 points, so there are (253)=2300\binom{25}{3} = 2300 possible triangles.

Note that the only way a triangle doesn't work is if all the 33 points are in a straight line.

There are 55 rows, 55 columns, and 22 long diagonals. Each of these 1212 lines have 55 points, which means they contribute 12(53)=1210=120 12 \cdot \binom{5}{3} = 12 \cdot 10 = 120 degenerate triangles.

There are also the diagonal lines with 44 points, such as the line from (1,2)(1,2) to (4,5).(4,5). There are 44 of these lines, so they have 4(43)=44=16 4 \cdot \binom{4}{3} = 4 \cdot 4 = 16 degenerate triangles.

Similarly, there are 44 diagonal lines with 33 points. These give us 41=44 \cdot 1 = 4 extra triangles that don't work.

Now, we have to look at the lines with slopes of 12,2,12,\dfrac{1}{2}, 2, -\dfrac{1}{2}, and 2.-2.

There are 33 such lines for each slope, and they all have 33 points on them. Therefore, they contribute 431=12 4 \cdot 3 \cdot 1 = 12 more triangles to discount.

The total number of working triangles is then 230012016412 2300 - 120 - 16 - 4 - 12 =2148.= 2148. Thus, B is the correct answer.

← Problem 22#22
Full Exam

Problem 23 in Other Years